$\begin{array}{l}+)\quad A_1 = -\dfrac{2}{\sqrt x +1}\\ Ta\,\,có:\quad\sqrt x \geq 0\\ \Rightarrow \sqrt x + 1\geq 1\\ \Rightarrow \dfrac{1}{\sqrt x + 1} \leq 1\\ \Rightarrow -\dfrac{2}{\sqrt x + 1} \geq -2\\ Hay \,\,A_1 \geq -2\\ \text{Dấu = xảy ra}\,\Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0\\ Vậy\,\,minA_1 = -2 \quad tại\,\,x=0\\+) \quad A_2 = \dfrac{\sqrt x - 1}{\sqrt x + 2} = 1 - \dfrac{3}{\sqrt x + 2}\\ Ta\,\,có:\,\sqrt x + 2 \geq 2\\ \Rightarrow \dfrac{3}{\sqrt x + 2} \leq \dfrac{3}{2}\\ \Rightarrow 1 - \dfrac{3}{\sqrt x + 2} \geq 1 - \dfrac{3}{2} = -\dfrac{1}{2}\\ Hay\,\,A_2 \geq -\dfrac{1}{2}\\ \text{Dấu = xảy ra}\,\Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0\\ Vậy\,\,minA_2 = -\dfrac{1}{2}\quad tại\,\,x=0\\+) \quad A_3 = \dfrac{x+5}{\sqrt x + 2}= \sqrt x +2 + \dfrac{9}{\sqrt x + 2} - 4\\ Ta\,\,có:\,\sqrt x +2 + \dfrac{9}{\sqrt x + 2} \geq 2\sqrt{(\sqrt x +2)\left(\dfrac{9}{\sqrt x + 2}\right)} = 6\\ \Rightarrow \sqrt x +2 + \dfrac{9}{\sqrt x + 2} - 4 \geq 6 - 4 = 2\\ Hay\,\,A_3\geq 2\\ \text{Dấu = xảy ra}\,\Leftrightarrow \sqrt x + 2 = \dfrac{9}{\sqrt x + 2}\\ \Leftrightarrow (\sqrt x + 2)^2 = 9\\ \Leftrightarrow \left[\begin{array}{l}\sqrt x + 2 = 3\\\sqrt x + 2 = -3\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =1\\\sqrt x + 2 = -3\quad (vô\,\,nghiệm)\end{array}\right.\\ Vậy\,\,minA_3 = 2\quad tại\,\,x = 1\\+) \quad A_4 = \dfrac{x + 3}{\sqrt x + 3} = \sqrt x + 3 + \dfrac{12}{\sqrt x + 3} - 6\\ Ta\,\,có:\,\sqrt x + 3 + \dfrac{12}{\sqrt x + 3} \geq 2\sqrt{(\sqrt x + 3)\left(\dfrac{12}{\sqrt x + 3}\right)}=4\sqrt3\\ \Rightarrow \sqrt x + 3 + \dfrac{12}{\sqrt x + 3} - 6 \geq 4\sqrt3 - 6\\ Hay \,\,A_4 \geq 4\sqrt3 - 6\\ \text{Dấu = xảy ra}\,\Leftrightarrow \sqrt x + 3 = \dfrac{12}{\sqrt x + 3}\\ \Leftrightarrow (\sqrt x + 3)^2 = 12\\ \Leftrightarrow \left[\begin{array}{l}\sqrt x + 3 = 2\sqrt3\\\sqrt x + 3 = -2\sqrt3\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = 21 - 12\sqrt3\\\sqrt x + 3 = -2\sqrt3\quad (vô\,\,nghiệm)\end{array}\right.\\ Vậy\,\,minA_4 = 4\sqrt3 - 6\quad tại\,\,x=21 - 12\sqrt3\\+) \quad B_1 = \dfrac{x+1}{\sqrt x} = \sqrt x + \dfrac{1}{\sqrt x}\\ Ta\,\,có:\,\sqrt x + \dfrac{1}{\sqrt x} \geq 2\sqrt{\sqrt x.\dfrac{1}{\sqrt x}} = 2\\ Hay\,\,B_1 \geq 2\\ \text{Dấu = xảy ra}\, \Leftrightarrow \sqrt x = \dfrac{1}{\sqrt x} \Leftrightarrow x = 1\\ Vậy\,\,minB_1 = 2 \quad tại\,\,x=1\\+) \quad B_2 = \dfrac{x + \sqrt x + 1}{\sqrt x} =\sqrt x + \dfrac{1}{\sqrt x} + 1\\ Ta\,\,có:\,\sqrt x + \dfrac{1}{\sqrt x} \geq 2\sqrt{\sqrt x.\dfrac{1}{\sqrt x}} = 2\\ \Rightarrow \sqrt x + \dfrac{1}{\sqrt x} +1 \geq 2 + 1 = 3\\ Hay\,\,B_2 \geq 3\\ \text{Dấu = xảy ra}\, \Leftrightarrow \sqrt x = \dfrac{1}{\sqrt x} \Leftrightarrow x = 1\\ Vậy\,\,minB_2 = 3 \quad tại\,\,x=1\\\end{array}$
